Solve x^2 - 10000x - 10000 = 0 without a calculator! A Precalculus Investigation

BACKGROUND/OVERVIEW
No, there's no mistake in the constant term in that equation. Imagine giving this to your Precalculus/Math Analysis/Adv Math students! Actually, 'solving' it with the TI-84 requires some effort using Solver since one needs to make an approximate guess or adjust the lower bound so that the positive root is obtained. Using the graph is no 'walk in the park' either! The TI-89 or Mathematica would have much less difficulty in displaying the exact radical form or a suitable decimal approximation but they may not be within reach. Perhaps an important issue here is that sometimes technology gives us unexpected or even inaccurate results. That's when students need some understanding of theory to recognize the limitations of the technology and adjust accordingly.

Here's the point of all this. The given quadratic is not factorable over the integers, however we can replace it with a 'nicer' quadratic that is. The roots of the desired quadratic can be shown to be approximately the same as the 'nice' quadratic and we can show that the absolute error is less than two ten-thousanths (and a much much smaller relative or % error)! Does this 'numerical analysis' have any practical value? Why approximate roots when powerful technology can produce exact answers? Do professionals who need to apply mathematics to the solution of 'real' problems ever use such approximation techniques? Could it be that theory actually provides practical application!

THE INVESTIGATION
(1) Show that the roots of the x²-10000x-10001 = 0 are 10001 and -1 by factoring.

(2) Show that the roots of x²-10000x-10000=0 can be approximated by 10001 and -1 with an error of less than 0.0001.
(a) By direct calculation: Using the quadratic formula and, yes, you may use the calculator!

(b) (Challenging) By comparing, in general,
(*) the roots of x²-bx-(b+1)=0 and
(**) the roots of x²-bx-b=0.Here we are assuming that b > 0.
(i) First show by factoring that the roots of equation (*) are b+1 and -1.
(ii) Then use the quadratic formula to express the roots of (**) in terms of b.
(iii) Compare the positive roots of these equations by subtracting them and (after algebraic manipulation and simplication), show that the absolute value of the difference is less than 1/(b+1).
Note: For b=10000, this error is therefore less than 0.0001.

(c) Explain intuitively why the roots of the original equation and the 'approximating' equation are virtually the 'same' for 'large' values of b. One possibility here is to consider how the graphs of the associated quadratic functions are related. What do they have in common? How are they different?
Note: Subtle point here for students. Even though the difference of the function values (i.e., y-values) is always 1, this is not true of the difference between their zeros! This may be the essence of the numerical analysis in this investigation.

EXTENSION/PRACTICE
Ok, now "solve" x² - (googol)x - googol = 0 without a calculator.

RELATED PROBLEM:
Without your calculator show that √(10001) - √(10000) is less than 0.005.
Does this provide us with an effective method of approximating the square root of some large numbers or is it limited and impractical?

For Calculus students: How does this compare to using linearization to approximate the square root?

For more advanced calculus students: Newton's Method? The Binomial Formula (using fractional exponents)? A Taylor Polynomial approximation? All equivalent?

But it's not in the Standards: Finding imaginary roots, completing the square, factoring and other 'obsolete' topics...

Remember the good old days when students solved 'quadratic-type' equations? Of course, many are still doing this but it is fast becoming a lost art (and some of you may feel it should be!). It is not required in any state math standards or Achieve's, so there's no reason to mention it, right?

Below you will find a 4th-degree (quartic) polynomial equation. The rational root theorem won't help because there are no rational roots. The graphing calculator won't help because there are no real roots! Ok, maybe Mathematica and other Symbolic algebra software could do this, but who exactly programmed this?

Using substitution to rewrite certain 4th degree equations as quadratics (so-called 'biquadratic' equations) used to be covered in some Algebra 2 or advanced classes. Some of you may feel nostalgic about this. However, our challenge today is to solve this by at least TWO 'radically' different methods and then show the solutions are equivalent!

Here's your equation:
x⁴ + 3x² + 4 = 0

(a) Explain, without solving, why this equation has no real roots. Should ALL students in Algebra 2 and beyond be able to answer this one?

(b) Solve, by substituting y for x² and using the quadratic formula. You should eventually arrive at 4 imaginary solutions. This is the way I was taught to solve it, eons ago.

(c) Solve by completing the square and factoring. [Definitely not the first method I would have thought of way back when...]

(d) Show your results are equivalent. This may be annoying! So, which method is easier in your opinion?

(e) Any other method for finding imaginary solutions?

QUICK OPINION POLL
(1) Completing the square (not to mention factoring) is no longer an important topic and should be deemphasized in our curriculum (or omitted).
By the way, is it explicitly mentioned in your state's math standards for Gr 8-12?
(YES NO)

(2) The equation in this post has little relevance to the 21st century and Dave should be ashamed for publishing such trash. Besides, this topic is not included in the Algebra 2 Standards developed by Achieve and ADP.
(YES NO)

You've perhaps assumed that since I've been discussing and complimenting Achieve's standards and the new Algebra 2 End of Course Exam, that I would no longer advocate exposing students to this kind of traditional mechanical 'exercise.'

Well, I taught from the AP Calculus syllabus and I still made time to discuss some ideas and methods that were not 'required'! Further, who exactly will be the ones left on this planet who know how to find imaginary roots for this type of polynomial equation that has no real roots! In case you're wondering, this kind of question has traditionally been taught in Asian countries and still is! (Dave, can you document that? Sure...)